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Electronic Journal of Differential Equations, Vol. 2014 (2014), No. 175, pp. 1–10. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu THREE-POINT THIRD-ORDER PROBLEMS WITH A SIGN-CHANGING NONLINEAR TERM JOHNNY HENDERSON, NICKOLAI KOSMATOV Abstract. In this article we study a well-known boundary value problem u000 (t) = f (t, u(t)), 0 0 < t < 1, 00 u(0) = u (1/2) = u (1) = 0. With u0 (η) = 0 in place of u0 (1/2) = 0, many authors studied the existence of positive solutions of both the positone problems with η ≥ 1/2 and the semipositone problems for η > 1/2. It is well-known that the standard method successfully applied to the semi-positone problem with η > 1/2 does not work for η = 1/2 in the same setting. We treat the latter as a problem with a signchanging term rather than a semi-positone problem. We apply Krasnosel’skiı̆’s fixed point theorem [4] to obtain positive solutions. 1. Introduction We study the third-order nonlinear boundary-value problem u000 (t) = f (t, u(t)), 0 0 < t < 1, 00 u(0) = u (1/2) = u (1) = 0. (1.1) (1.2) with a sign-changing nonlinearity. Equation (1.1) satisfying the three-point condition u(0) = u0 (η) = u00 (1) = 0, (1.3) with η ≥ 1/2 has been studied by many authors [2, 7, 11]. We mention also relevant results in [1, 3], where, under nonlocal conditions involving Stieltjes integrals, the positone case was considered. A good theory of positive solutions for semi-positone problems with η > 1/2 is developed in [5, 8, 9, 10] (and the references therein). In particular, Yao [8] obtained a positive solution of the boundary value problem similar to (1.1), (1.3). The author assumed that the function f : [0, 1] × R+ → R satisfies the Carathéodory conditions and there exists a nonnegative function h ∈ L1 [0, 1] such that f (t, u) ≥ h(t), (t, u) ∈ [0, 1] × R+ . Our paper is motivated by [8] where, we believe, the idea of a non-constant lower bound −h(t) for the inhomogeneous term was originally used for the boundary value problem similar to (1.1), (1.2). The author refers to this type of problem as weakly semipositone. 2000 Mathematics Subject Classification. 34B15, 34B16, 34B18. Key words and phrases. Green’s function; fixed point theorem; positive solutions; third-order boundary-value problem. c 2014 Texas State University - San Marcos. Submitted June 10, 2014. Published August 14, 2014. 1 2 J. HENDERSON, N. KOSMATOV EJDE-2014/175 Prior to [8], similar semipositone problems have been solved effectively [10] only for f : [0, 1] × R → [−M, ∞) due to selection of h ≡ M > 0. As in [8], in this paper, we need only f : [0, 1] × R+ → R. Regardless of the choice of h, as a first step, one translates the semipositone problem into a positone problem using the transformations u 7→ v − u0 and f (·, u) 7→ f (·, v − u0 ) + h(·), where u0 is a unique solution of the problem with the nonlinear term replaced with h. Subsequently, the positone problem is converted into an integral equation, which is shown to have one or, depending on conditions of f , several positive solutions. Finally, an important feature of this approach is that it requires the inequality v(t) ≥ u0 (t) to hold for a fixed point of the corresponding integral operator. This comparison depends on the properties of Green’s function, or in particular, on the function appearing in the definition of a cone, and the solution u0 . The case of η = 1/2 stands alone since this type of approach used by many authors to study the case η > 1/2 does not readily apply to the case η = 1/2. The difficulty arises when we attempt to obtain the inequality v(t) ≥ u0 (t) for η = 1/2. Since problem (1.1), (1.2) cannot be treated as a semipositone problem, we adopt a new set of assumptions and consider a sign-changing nonlinearity. We are unaware of any results on the case η = 1/2 with a sign-changing nonlinear term. Another benefit is that we can also obtain new results for the case η > 1/2 with a signchanging nonlinearity by employing the concept of a sign-changing lower bound g0 . We think that it would not be difficult to extend our results to the case of f satisfying the Carathéodory conditions and even treat singularities as in [10]. Here we settle for a continuous sign-changing nonlinear term. 2. Properties of Green’s function Let g0 ∈ C[0, 1]. Then the differential equation u000 (t) = g0 (t), 0 < t < 1, (2.1) satisfying the boundary condition (1.2) has a unique solution Z Z t2 1 1 t 2 (t − s) g0 (s) ds − g0 (s) ds u0 (t) = 2 0 2 0 Z 1/2 1 Z 1 1 g0 (s) ds − − s g0 (s) ds . +t 2 0 2 0 Using Green’s function 1 1 1 G(t, s) = (t − s)2 χ[0,t] (s) + (t − t2 ) − t − s χ[0,1/2] (s), 2 2 2 for (t, s) ∈ [0, 1] × [0, 1], we have Z 1 u0 (t) = G(t, s)g0 (s) ds. (2.2) 0 Let s2 1 χ[0,1/2] (s) + χ[1/2,1] (s), s ∈ [0, 1]. 2 8 We revisit the important properties [5] of (2.2) used in cone-theoretic methods: G0 (s) = G(1/2, s) = q(t)G0 (s) ≤ G(t, s) ≤ G0 (s), (t, s) ∈ [0, 1] × [0, 1], (2.3) EJDE-2014/175 THREE-POINT THIRD-ORDER PROBLEMS 3 where q(t) = 4(t − t2 ). (2.4) Also, Z t∈[0,1] 1 G(t, s) ds = L = max 0 1 , 12 (2.5) and, for 0 < α < 1/2, 1−α Z 1 (2 − 4α3 − 3α). 24 G0 (s) ds = C= α (2.6) Note that if Z 1 g0 (t) dt ≥ 0, t ∈ [0, 1], (2.7) t then u0 (t) is concave in [0, 1]. If, in addition, Z Z 1/2 1 1 1 2 u0 (1) = − s g0 (s) ds ≥ 0, (1 − s) g0 (s) ds − 2 0 2 0 (2.8) then u0 (t) ≥ 0. Note that neither (2.7) nor (2.8) requires g0 (t) ≥ 0 in all of [0, 1]. Moreover, if g0 (t) ≥ 0 in [0, 1] and g0 (t) > 0 in some [α, β] ⊂ [0, 1], then u0 (1) > 0. This represents a difficulty due to the fact that one can not achieve the inequality q(t) ≥ µu0 (t) in [0, 1] for any µ > 0 (as q(1) = 0 while u0 (1) > 0). For this reason, the case η = 1/2 is forbidden in approaching (1.1), (1.3) as a semipositone problem. If the identity takes place in (2.8), that is, u0 (1) = 0 is enforced, then we are in position to compare q(t) and u0 (t) in the next lemma. Lemma 2.1. Let g0 ∈ C[0, 1] satisfy (2.7) and suppose that the identity holds in (2.8). Then there exists a constant µ > 0 such that q(t) ≥ µu0 (t), t ∈ [0, 1]. (2.9) Proof. Since the function q − µu0 vanishes at the end-points of [0, 1], it suffices to obtain µ > 0 such that −q 00 (t) ≥ −µu000 (t) in [0, 1]. That is, Z 1 8≥µ g0 (s) ds, t ∈ [0, 1]. t By (2.7), there exists 0 < τ < 1 and µ > 0 such that Z 1 Z 1 µ g0 (s) ds ≤ µ g0 (s) ds = 8. t (2.10) τ Suppose that the function f in (1.1) satisfies (H1) f ∈ C([0, 1] × R+ , R); (H2) there exists a function g0 ∈ C[0, 1] such that (a) f (t, z) + g0 (t) ≥ 0 in [0, 1] × R+ ; R1 (b) for all t ∈ [0, 1], t g0 (s) ds ≥ 0; (c) Z Z 1/2 1 1 1 2 (1 − s) g0 (s) ds − − s g0 (s) ds = 0. 2 0 2 0 4 J. HENDERSON, N. KOSMATOV EJDE-2014/175 Remark 2.2. It is easy to find a function g0 satisfying (H2) (b) and (c). For example, one can take g0 (t) = a(2t − 1), a > 0. Of course, an example of f (t, z) that fits (H2) (b) is also easy to obtain. Remark 2.3. If the inequality (2.8) is replaced with the strict inequality, we cannot expect Lemma 2.1 to hold. So, in this paper, we need the identity in (H2) (c). If, instead of u0 (1/2) = 0, we impose u0 (η) = 0 with η > 1/2, then the problem (2.1), (1.3) has a unique solution Z Z Z η Z 1 1 t t2 1 u0 (t) = (t−s)2 g0 (s) ds− g0 (s) ds+t η g0 (s) ds− (η −s)g0 (s) ds . 2 0 2 0 0 0 Again, the assumption (H1) (b) guarantees that u0 is concave in [0, 1]. So, if Z Z Z η 1 1 1 1 2 (1 − s) g0 (s) ds + η − g0 (s) ds − (η − s)g0 (s) ds ≥ 0, u0 (1) = 2 0 2 0 0 then u(t) ≥ 0 in [0, 1]. Similarly, the analogue of q(t), in this case [5], is 1 p(t) = 2 (2ηt − t2 ). η Noting that p(1) 6= 0 and u0 , p are concave concave in [0, 1], we can easily obtain an analogue of Lemma 2.1 asserting the existence of µ > 0 such that p(t) ≥ µu0 (t) in [0, 1]. This would give a more general result than in [10, Lemma 2.1 (4)], which is derived for g0 ≡ M > 0. This would also allow us to extend the results of [8] concerning an analogue of (1.1), (1.3), where h(t), which serves the purpose of g0 (t), is assumed to be nonnegative. We modify the problem (1.1), (1.2) as follows. First, we define ( f (t, z) + g0 (t), (t, z) ∈ [0, 1] × [0, ∞), fp (t, z) = f (t, 0) + g0 (t), (t, z) ∈ [0, 1] × (−∞, 0). Next, we consider the equation v 000 (t) = fp (t, v(t) − u0 (t)), t ∈ (0, 1), (2.11) under the boundary conditions (1.2). We can easily obtain the next lemma. Lemma 2.4. The function u is a positive solution of the boundary value problem (1.1), (1.2) if, and only if, the function v = u + u0 is a solution of the boundary value problem (2.11), (1.2) satisfying v(t) ≥ u0 (t) in [0, 1]. In the Banach space B = C[0, 1] endowed with usual max-norm, we consider the operator Z 1 T v(t) = G(t, s)fp (s, v(s) − u0 (s)) ds, (2.12) 0 where G(t, s) is given by (2.2). By (H1), T : B → B is completely continuous. Using the function q defined by (2.4), we introduce the cone C = {v ∈ B : v(t) ≥ q(t)kvk, t ∈ [0, 1]}. By (2.3), T : C → C and it is also easy to show that a fixed point of T is a solution of (2.11), (1.2). In particular, v(t) ≥ γkvk, 2 t ∈ [τ, 1 − τ ], (2.13) where γ = mint∈[α,1−α] q(t) = 4(α − α ), and κ = maxt∈[α,1−α] q(t) = q(1/2) = 1. EJDE-2014/175 THREE-POINT THIRD-ORDER PROBLEMS 5 Theorem 2.5 ([4]). Let B be a Banach space and let C ⊂ B be a cone in B. Assume that Ω1 , Ω2 are open with 0 ∈ Ω1 , Ω1 ⊂ Ω2 , and let T : C ∩ (Ω2 \ Ω1 ) → C be a completely continuous operator such that either (i) kT uk ≤ kuk, u ∈ C ∩ ∂Ω1 , and kT uk ≥ kuk, u ∈ C ∩ ∂Ω2 , or; (ii) kT uk ≥ kuk, u ∈ C ∩ ∂Ω1 , and kT uk ≤ kuk, u ∈ C ∩ ∂Ω2 . Then T has a fixed point in C ∩ (Ω2 \ Ω1 ). 3. Positive solutions To use Theorem 2.5, following [8] we introduce the “height” functions φ, ψ : R+ → R+ defined by φ(r) = max{f (t, z − u0 (t)) + g0 (t) : t ∈ [0, 1], z ∈ [0, r]}, ψ(r) = min{f (t, z − u0 (t)) + g0 (t) : t ∈ [α, 1 − α], z ∈ [γr, r]}. Now we present our main results. Theorem 3.1. Assume that (H1) and (H2) hold. Suppose that there exist r, R > 0 such that µ1 < r < R, where µ > 0 satisfies (2.9), (2.10), and (H3) φ(r) ≤ 12r and ψ(R) ≥ 24R 2−4α3 −3α . Then the boundary-value problem (1.1), (1.2) has at least one positive solution. Proof. Let Ω1 = {v ∈ B : kvk < r}, Ω2 = {v ∈ B : kvk < R}. For u ∈ C ∩ ∂Ω1 , we have v(s) − u0 (s) ≥ q(s)kvk − u0 (s) ≥ (µr − 1)u0 (s) ≥ 0, s ∈ [0, 1]. This implies that fp (s, v(s) − u0 (s)) = f (s, v(s) − u0 (s)) + g0 (s), s ∈ [0, 1]. In particular, f (s, v(s) − u0 (s)) + g0 (s) ≤ φ(r), s ∈ [0, 1], 0 ≤ v(s) ≤ r. Thus, by (2.5) and (H3), Z 1 kT vk = max t∈[0,1] G(t, s)fp (s, v(s) − u0 (s)) ds 0 Z 1 ≤ max t∈[0,1] G(t, s) ds φ(r) 0 1 φ(r) ≤ r. 12 That is, kT vk ≤ kvk for all v ∈ C ∩ ∂Ω1 . Let v ∈ C ∩ ∂Ω2 . Since R > r, we have v(s) − u0 (s) ≥ (µR − 1)u0 (s) ≥ 0, s ∈ [0, 1]. Then, for all s ∈ [α, 1 − α], we have, recalling (2.13), = Lφ(r) = R ≥ v(s) ≥ q(s)kvk ≥ γR. Hence fp (s, v(s) − u0 (s)) = f (s, v(s) − u0 (s)) + g0 (s) ≥ ψ(R), 6 J. HENDERSON, N. KOSMATOV EJDE-2014/175 for s ∈ [α, 1 − α], γR ≤ v(s) ≤ R. Then, by (2.6) and (H3), Z 1 G(t, s)fp (s, v(s) − u0 (s)) ds kT vk = max t∈[0,1] 0 Z 1−α G(t, s)fp (s, v(s) − u0 (s)) ds ≥ max t∈[0,1] α Z 1−α ≥ max t∈[0,1] q(t)G0 (s) ds ψ(R) α Z 1−α = max q(t) t∈[0,1] G0 (s) ds ψ(R) α = κCψ(R) ≥ R. That is, kT vk ≥ kvk for all v ∈ C ∩ ∂Ω2 . By Theorem 2.5, there exists v0 ∈ C with u(t) = v0 (t)−u0 (t) ≥ (µr −1)u0 (t) ≥ 0 in [0, 1]. By Lemma 2.4, u is a positive solution of the sign-changing problem (1.1), (1.2). Now we give an example of the right side of (1.1) satisfying the assumptions of Theorem 3.1. Example. Let f (t, z) = 6z 2 +32(1−2t) for z ≥ 0, t ∈ [0, 1]. Then f (t, z)+g0 (t) ≥ 0 with g0 (t) = 32(2t − 1). Of course, (H1) and (H2) hold and Z 1 Z 1 g0 (t) ds ≤ g0 (t) ds = 8. t 1/2 Hence we can choose µ = 1 and note that µr > 1, if we choose r = 2. Then, recalling that v(s) − u0 (s) ≥ 0, f (t, v(s) − u0 (s)) + g0 (t) = 6(v(s) − u0 (s))2 ≤ 6kvk2 = 24 = 12r, for all v ∈ C ∩ ∂Ω1 . This shows that the first condition of (H3) is fulfilled. It is easy to see that Z 1 Z 1 8 ku0 k ≤ max G(t, s)|g0 (s)| ds ≤ max G(t, s) ds kg0 k = . 3 t∈[0,1] 0 t∈[0,1] 0 Let now α = 1/4 so that C = 19/384 and γ = 4(α − α2 ) = 3/4. Then, for all s ∈ [1/4, 3/4], v ∈ C ∩ ∂Ω2 , where R = 13, we have f (t, v(s) − u0 (s)) + g0 (t) = 6(v(s) − u0 (s))2 ≥ 6(γkvk − ku0 k)2 7225 39 8 2 − = =6 4 3 24 4992 24R > = . 19 2 − 4α3 − 3α The above shows that the second part of (H3 ) is also verified. Hence a solution v0 exists in the cone and 2 ≤ kv0 k ≤ 13. The next result can be shown along the similar lines. Theorem 3.2. Assume that (H1) and (H2) hold. Suppose that there exist r, R > 0 such that µ1 < r < R, where µ > 0 satisfies (2.9), (2.10), and EJDE-2014/175 THREE-POINT THIRD-ORDER PROBLEMS (H4) φ(R) ≤ 12R and ψ(r) ≥ 7 24r 2−4α3 −3α . Then the boundary0value problem (1.1), (1.2) has at least one positive solution. In conclusion of this paper presents a multiplicity result for (1.1), (1.2) which now is considered as a nonlinear eigenvalue problem. That is, u000 (t) = λf (t, u(t)), 0 < t < 1, (3.1) subject to (1.2). The result including the assumptions and the method of proof echoes that of Ma [6], where a fourth order semipositone boundary-value problem with dependence on the first derivative was studied. The presence of the parameter λ > 0 provides an additional control on the growth of the right side. We introduce a new set of assumptions as follows: (M1) there exists an interval [α, 1 − α] ⊂ (0, 1) such that lim u→∞ f (t, u) = ∞, u uniformly in [α, 1 − α]; (M2) f (t, 0) > 0, t ∈ [0, 1]. Our next result is a multiplicity criterion. Theorem 3.3. Assume that (H1), (H2), (M1), (M2) hold. Then the boundaryvalue problem (3.1), (1.2) has at least two positive solutions provided λ > 0 is small enough. Proof. We will construct open nonempty subsets Ωi = {v ∈ C : kvk = Ri }, i = 1, . . . , 4. Now, we consider the operator Z 1 T v(t) = λ G(t, s)fp (s, v(s) − λu0 (s)) ds, 0 where u0 is the solution of u000 = g0 subject to (1.2) and fp as above. Let the R1 > 0. Then Z 1 kT vk = max λ G(t, s)fp (s, v(s) − λu0 (s)) ds ≤ λLφ(R1 ) ≤ R1 t∈[0,1] 0 for all v ∈ C ∩ ∂Ω1 , provided Lφ(R1 ) . R1 Let v ∈ C ∩ ∂Ω2 , where R2 > R1 . Then, by Lemma 2.1 with Z 1 µ max g0 (s) ds = 8. λ≤ t∈[0,1] (3.2) t Note that the equation in (M2) holds with fp in place of f . Thus given A > 0, there exists h ≥ γ2 R2 such that fp (t, z) > Az for all z ≥ h and t ∈ [α, 1 − α]. For every λ in (3.2), there exists a constant A > 0 such that 1 λCγA ≥ 1, 2 where C is given by by (2.6). For all s ∈ [α, 1 − α], we have v(s) − λu0 (s) ≥ v(s) − λ λ 1 γ q(s) = v(s) − v(s) ≥ v(s) ≥ R2 µ µR2 2 2 (3.3) 8 J. HENDERSON, N. KOSMATOV EJDE-2014/175 provided λ≤ µR2 . 2 (3.4) Hence fp (s, v(s) − λu0 (s)) ≥ A(v(s) − λu0 (s)) ≥ γA R2 , 2 s ∈ [α, 1 − α]. Then, by (3.3)), and recalling that κ = 1, Z 1 kT vk = max λ G(t, s)fp (s, v(s) − λu0 (s)) ds t∈[0,1] 0 Z 1−α ≥ λ max t∈[0,1] q(t)G0 (s) ds α Z 1−α G0 (s) ds = λ max q(t) t∈[0,1] α γA R2 2 γA R2 2 γA = λκC R2 ≥ R2 . 2 That is, kT vk ≥ kvk for all v ∈ C ∩ ∂Ω2 . As in Theorem 3.1, we have a solution v1 such that R1 ≤ kv1 k ≤ R2 for every 0 < λ ≤ λ0 = min µR2 R1 , . Lφ(R1 ) 2 To make use of the assumption (M2 ), we note that there exist a, b > 0 such that f (t, z) ≥ b for all t ∈ [0, 1] and z ∈ [0, a] and introduce a “truncation” of f given by ( f (t, z), (t, z) ∈ [0, 1] × [0, a]), ft (t, z) = f (t, a), (t, z) ∈ [0, 1] × (a, ∞). Consider now u000 (t) = λft (t, u(t)), 0 < t < 1, (3.5) subject to (1.2). The operator, whose fixed point will be shown to be (a second) solution of (1.1), (1.2), is Z 1 T v(s) = λ G(t, s)ft (s, v(s)) ds. 0 Choose R3 < min{R1 , a}. Then, as in the proof of Theorem 3.1, kT vk ≤ λLφ(R3 ), where φ(R3 ) = max{f (t, z) : t ∈ [0, 1], z ∈ [0, R3 ]}. So, if λ < min R3 , λ0 , Lφ(R3 ) then kT vk ≤ kvk for all v ∈ C ∩ ∂Ω3 . Choose λ according to (3.6). Since lim z→0+ b ft (t, z) ≥ lim =∞ + z z→0 z uniformly in [0, 1]. Hence there exists 0 < R4 < R3 such that ft (t, z) ≥ Bz, t ∈ [0, 1], z ∈ [0, R4 ], (3.6) EJDE-2014/175 THREE-POINT THIRD-ORDER PROBLEMS 9 where Z λBD ≥ 1, t∈[0,1] 1 G(t, s)q(s) ds. D = max 0 Then, for all v ∈ C ∩ ∂Ω4 , 1 Z kT vk = max λ t∈[0,1] G(t, s)ft (s, v(s)) ds 0 Z 1 ≥ max λB t∈[0,1] G(t, s)v(s) ds 0 Z ≥ λB max t∈[0,1] 1 G(t, s)q(s)R4 ds 0 = λBDR4 ≥ kvk. Thus, there exists a positive solution v2 with R4 ≤ kv2 k ≤ R3 for every λ > 0 satisfying (3.6). Finally, since R3 < R1 , the solutions are distinct. Acknowledgments. The article was prepared while the second author was on sabbatical leave at Baylor University. He is grateful to Baylor University for its support and hospitality. References [1] J. R. Graef, J. R. L. Webb; Third order boundary value problems with nonlocal boundary conditions, Nonlinear Anal. 71 (2009), 15421551. [2] J. R. Graef, B. Yang; Positive solutions of a nonlinear third order eigenvalue problem, Dynam. Systems Appl. 15 (2006), no. 1, 97–110. [3] G. Infante, P. Pietramala; A third order BVP subject to nonlinear boundary conditions, Math. Bohem. 135 (2010), 113121. [4] M. A. Krasnosel’skiı̆; “Topological Methods in the Theory of Nonlinear Integral Equations”, (English) Translated by A.H. Armstrong; A Pergamon Press Book, MacMillan, New York, 1964. [5] K. Q. Lan; Eigenvalues of semi-positone Hammerstein integral equations and applications to boundary value problems, Nonlinear Anal. 71 (2009), 5979-5993. [6] R. Ma; Multiple positive solutions for a semipositone fourth-order problem, Hiroshima Math. J. 33 (2013), 217-227. [7] Y. Sun; Existence of triple positive solutions for a third-order three-point boundary-value problem, J. Comp. Appl. Math. 221 (2008), 194–201. [8] Q. Yao; Positive solutions of a weak semipositone third-order three-point boundary value problem, J. Math. Res. Exp. 30 (2010), no. 1, 173–180. [9] Q. Yao; Positive solutions of singular third-order three-point boundary value problems, J. Math. Anal. Appl. 354 (2009), 207–212. [10] H. Yu, L. Haiyan, Y. Liu, Multiple positive solutions to third-order three-point singular semipositone boundary value problem, Proc. Indian Acad. Sci. (Math. Sci.) 114 (2004), no. 4, 409–422. [11] L. Zhang, B. Sun, C. Xing; Existence of solutions to a third-order three-point boundary value problem, Ann. Differential Equations 28 (2012), no. 3, 368–372. Johnny Henderson Department of Mathematics, Baylor University, Waco, TX 76798-7328, USA E-mail address: Johnny [email protected] 10 J. HENDERSON, N. KOSMATOV EJDE-2014/175 Nickolai Kosmatov Department of Mathematics and Statistics, University of Arkansas at Little Rock, Little Rock, AR 72204-1099, USA E-mail address: [email protected]